Integral of ln(1+x^2) (by parts)

✍🏼 integralsforyou.com - Integral of ln(1+x^2) - How to integrate it step by step using integration by parts!
✅ 𝐃𝐞𝐫𝐢𝐯𝐚𝐭𝐞 𝐭𝐨 𝐜𝐡𝐞𝐜𝐤 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Derivative of x*ln(1+x^2) - 2x + 2arctan(x) = • Derivative of x*ln(1+x...
🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫:
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[@IntegralsForYou]

🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫: ► Integral searcher 👉integralsforyou.com/integral-searcher 🎓 𝐇𝐚𝐯𝐞 𝐲𝐨𝐮 𝐣𝐮𝐬𝐭 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 𝐚𝐧 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝? 𝐅𝐢𝐧𝐝 𝐞𝐚𝐬𝐲, 𝐦𝐞𝐝𝐢𝐮𝐦 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡 𝐥𝐞𝐯𝐞𝐥 𝐞𝐱𝐚𝐦𝐩𝐥𝐞𝐬 𝐡𝐞𝐫𝐞: ► Integration by parts 👉integralsforyou.com/integration-methods/integration-by-parts ► Integration by substitution 👉integralsforyou.com/integration-methods/integration-by-substitution ► Integration by trig substitution 👉integralsforyou.com/integration-methods/integration-by-trig-substitution ► Integration by Weierstrass substitution 👉integralsforyou.com/integration-methods/integration-by-weierstrass-substitution ► Integration by partial fraction decomposition 👉integralsforyou.com/integration-methods/integration-by-partial-fraction-decomposition 👋 𝐅𝐨𝐥𝐥𝐨𝐰 @𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬𝐟𝐨𝐫𝐲𝐨𝐮 𝐟𝐨𝐫 𝐚 𝐝𝐚𝐢𝐥𝐲 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥! 😉 📸 instagram.com/integralsforyou/

[@arantzasepulveda1771]

I was stuck in this integral for hours thank you so much for uploading the solving process

[@IntegralsForYou]

My pleasure! Welcome and enjoy the channel! 😊😊

[@laasd1613]

On behalf of all your viewers, thank you. I can't imagine how difficult coming up with some of these solutions.

[@IntegralsForYou]

Thank you very much! It was a little bit hard when starting this channel, but after lots of integrals it become a lot more easier hehe

[@natalieyabalwashi3916]

we would appreciate if u did some explaining aswell..

[@jdubb5394]

Thank you sir! Saved my butt a couple times already 👍

[@IntegralsForYou]

And I am very happy to save it! 😊😜👍

[@atraps7882]

ahhh i was struggling trying to evaluate the second integral... the x^2 + 1 - 1 trick really helps!

[@IntegralsForYou]

Yeah, I love this trick! It avoids us to do a long polynomial division D=dq+r ==> D/d = q + r/d D = x^2 d = x^2+1 q = 1 r = -1 D = dq + r ==> x^2 = (x^2+1)*1 - 1 D/d = q + r/d ==> x^2/(x^2+1) = 1 - 1/(x^2+1)

[@Gatonaranja__]

Estuve buscando videos en español por horas para este ejercicio en específico, no se si entiendes pero gracias

[@IntegralsForYou]

Hola! Sí que entiendo sí! De hecho, cuando creé este canal dudé de si hacerlo en inglés o en español hasta que entendí que no era necesario ningún idioma si escribía el máximo de detalles posibles durante el video 💪 Gracias por tu comentario, espero que disfrutes del canal! Si quieres colaborar con una pequeña ayudita, suscríbete y te lo agradeceré aún más!❤

[@mzparvar5480]

It was very useful. Thank you, please add more video quality and sound.

[@IntegralsForYou]

I will! In fact, newer videos have better quality but still not sound...

[@theanonimo26able]

Thank you! You're amazing!!!!!!!

[@IntegralsForYou]

You're welcome! Thanks for your comment! 😉

[@OvnYf2307]

Gracias, ya me estaba volviendo loco el como resolver esta integrada, Like 1+

[@IntegralsForYou]

Me alegro de que te haya servido de ayuda! 😊Un saludo!

[@alinarikable]

Thank you so much!!!

[@IntegralsForYou]

You're welcome! ;-D

[@DaniGH96]

lol man love how u write, awesome, keep on doing this amazing job!!!

[@IntegralsForYou]

Thank you :-D

[@orhilel9945]

thank u , i would like to know how did u decide what will be dv and v as well

[@IntegralsForYou]

en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule 😉

[@dayhermarchan4568]

Gracias!

[@IntegralsForYou]

Un placer! :-D

[@MrFKD]

Thanks man that was helpful

[@IntegralsForYou]

My pleasure! ☺

[@nvfpv]

THERE IS A GOD THANK YOU SO MUCH

[@IntegralsForYou]

;-D You're welcome! ;-D

[@user-jz8fp3su7c]

amazing!

[@IntegralsForYou]

Thank you! 😊

[@ArantzaNava-zr9md]

Thank you for save me!!

[@IntegralsForYou]

My pleasure! 💪

[@redblack9103]

How do you integrate it with the substitution method? Can u please make a video or explain? Thanks

[@IntegralsForYou]

Hi Red Black, I am not sure we can do this integral by substitution, the two possible substitutions are u=1+x^2 or x=tan(u) and they make the integral more complicated

[@agv748]

GOOOOD 👍

[@IntegralsForYou]

Thaaaaanks! 😉😜

[@basicallytrying7856]

Help me a lot

[@IntegralsForYou]

Happy to know it! 😊😊

[@leonrique]

Is that thing about +1-1 a regular and common way to solve it, or there are easier ways?

[@IntegralsForYou]

Hi Leonrique Pereira! This is the easiest way to do it (when you get use to it :D). We use it to avoid doing a polynomial division D=dq+r: x^2 = 1*(1+x^2) - 1 ==> x^2/(1+x^2) = (1+x^2)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2) You will find some examples of polynomial division with (1+x^2) in this playlist: kzbin.info/aero/PLpfQkODxXi49YODSGByJH2O7dVRgzPKe2

[@larisapaula2233]

thank u, save my life. ^^

[@sunitakanav8524]

Thank you sir ji

[@IntegralsForYou]

My pleasure! 😊

[@mdjwy]

great!

[@IntegralsForYou]

;-D

[@user-nd7th3hy4l]

J= xln(1+x^2)-2x +2arctanx +k

[@IntegralsForYou]

Correct! 💪

[@2thecartoonsimp1]

Crazy notations

[@filpino234]

Hi, I have a quick question. Why do you add a "+1 -1" to the x^2 @ 1:47?

[@IntegralsForYou]

Hi! It is a little trick that is used when we have polynomials with the same degree in the numerator and in the denominator in order to avoid doing the long polynomial division. The trick consists of completing the numerator with +something-something to get the same numerator and denominator, which is equal to 1, and a new integral where the degree of the denominator is higher than the degree of the numerator. Let me write another example of the trick here: Integral of (x-1)/(x+2) dx = = Integral of (x-1+2-2)/(x+2) dx = = Integral of (x+2-1-2)/(x+2) dx = = Integral of (x+2-3)/(x+2) dx = = Integral of [(x+2)/(x+2) - 3/(x+2)] dx = = Integral of [1 - 3/(x+2)] dx = = Integral of 1 dx - 3*Integral of 1/(x+2) dx = = x - 3*ln|x+2| + C Hope it helped! 💪

[@Abhilashaisgood]

Appreciate your video, but i wonder why some hold their pen so weirdly. 🤔

[@IntegralsForYou]

Hehe thank you! Honestly, I don't know why I hold it that way... 😅

[@manueldavid84]

Thanks

[@IntegralsForYou]

My pleasure! ❤

[@big_snake431]

my hero

[@IntegralsForYou]

Thanks! 😊

[@AlxCass]

profe buen video solo que no entendi el por que pone (+1-1) gracias me esta ayudando :3

[@IntegralsForYou]

Hola Alejandro! el +1-1 sirve para modificar la integral sin variar el resultado ya que +1-1=0. Esta modificación la hacemos porque la integral de x^2/(1+x^2), que en teoría se resolvería mediante la división de polinomios, hacemos este pequeño truco que se hace en un momento y nos convierte x^2/(1+x^2) = (x^2+1-1)/(x^2+1) = (x^2+1)/(x^2+1) - 1/(x^2+1) = 1 - 1/(1+x^2). Para la integral de x^2/(1+x^2) no sé decirte el resultado a simple vista. Para la integral de 1 - 1/(1+x^2) sí, y es x - arctan(x) Espero haberte ayudado! (y si no, vuelve a preguntar ;-D)

[@a123467985]

1:45 awesome

[@IntegralsForYou]

😊Thanks! 😊

[@kuhaku3188]

Y no se puede hacer esta integral cambiando x^2 + 1 = t ??

[@IntegralsForYou]

No lo sé, puede que sí, pero se hace tan eterno que no merece la pena saberlo... Se hace largo por culpa de la expresión de dx en función de dt: t = 1+x^2 ==> t-1 = x^2 ==> sqrt(1-t) = x dt = 2x dx ==> dt = 2sqrt(1-t) dx ==> 1/2sqrt(1-t) dt = dx Integral de ln(1+x^2) dx = Integral de ln(t)*1/2sqrt(1-t) dt = ...

[@Xdetonando]

I dont get it, why the resolution isnt xln (1+x^2)-2x+k? What am i doing wrong?

[@IntegralsForYou]

Hi Xdetonando! Could you please write the steps you did? Thanks!

[@Xdetonando]

@@IntegralsForYou First i declared u as ln (1+x^2), du= 2x/1+x^2 dx, v = x, DV= dx, then used the formula, the result Was xln (1+x^2) - integral of x.(2x/1+x^2) dx, then i used division to null the x above, in the final i Had the integral of 2 dx, resulting in xln (1+x^2)-2x+C, dont know why this is wrong.

[@IntegralsForYou]

Hi, I don't know the exact mistake you did, but if I am not wrong, you are saying that 2x^2/(1+x^2) = 2 and that's impossible. When you have the integral of 2x^2/(1+x^2) you can do: A. As I do in the video: Integral of 2x^2/(1+x^2) dx = = 2*Integral of x^2/(1+x^2) dx = = 2*Integral of (x^2+1-1)/(1+x^2) dx = = 2*Integral of [ (x^2+1)/(1+x^2) - 1/(1+x^2) ] dx = = 2*Integral of [ 1 - 1/(1+x^2) ] dx = = 2*Integral of 1 dx - Integral of 1/(1+x^2) dx = = 2x - arctan(x) B. Polynomial division: Integral of 2x^2/(1+x^2) dx = = 2*Integral of x^2/(1+x^2) dx = D = dq + r x^2 = (1+x^2)*1 - 1 ==> x^2/(1+x^2) = (1+x^2)*1/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2) I don't know if I helped, but I am sure you did something wrong in your division.

[@nattpulgagzz5771]

¿Porque en el resultado sale arctan(x) ?

[@IntegralsForYou]

Hola NG glz! Si vas a ver las tablas de derivadas, verás que la derivada de arctan(x) es 1/(1+x^2). Por lo tanto, la integral de 1/(1+x^2) es directa y da como solución arctan(x). No sé si esta es la respuesta que buscabas... si no es así, vuélveme a preguntar detallándome un poco más tu duda ;-D

[@abdullahmahi4490]

how can i understand what i have to consider a s U or V

[@IntegralsForYou]

Hi! There is a rule called ILATE that is very useful. This video is a special case, but in general we have two functions multiplying. If you have time, I would recommend try u=first and dv=second and if it doesn't work, then try u=second and dv=first. When you learn it like this, then for the next integrals you will see which is u and dv at first sight. If you are in an exam and you don't have a lot of time, use the ILATE rule 😉

[@sergiohenrique1001]

god!

[@IntegralsForYou]

;-D

[@mintystorm43]

can you use trig sub ?

[@IntegralsForYou]

Hi Andrew Nicholas! You can try but by the moment I don't know how to continue after 2 steps... : Substitution: x=tan(u) dx= 1/cos^2(u) du Integral of ln(1+x^2) dx = = Integral of ln(1+tan^2(u)) 1/cos^2(u) du = = Integral of ln(1/cos^2(u)) 1/cos^2(u) du = ...

[@prismoid00]

You can use trig sub but the amount of steps are about the same. Though I do think the process is more elegant. x = tan(t) dx = sec^2(t)dt = { ln(1+ tan^2t)*sec^2t dt = { ln(sec^2t)*sec^2t dt = { 2ln(sec t)*sec^2t dt = 2{ ln(sec t)*sec^2t dt u = ln(sec t) du = sec t * tan t / sec t dt = tan t dt dv = sec^2t dt v = tan t = 2(ln(sec t)*tan t - {tan^2t dt) = 2(ln(sec t)*tan t - {(sec^2t - 1)dt) = 2(ln(sec t)*tan t - tan t + t) t = arctan x sec t = √(1 + tan^2t) = √(1 + x^2) = 2ln√(1+x^2)*x - 2x + 2arctanx = 2*1/2 ln(1+x^2)*x - 2(x - arctanx) = x*ln(1+x^2) - 2(x - arctanx) + c

[@LUISGONZALEZ-ir5pg]

De donde salio el numero uno que pones despues de l resultado primero? Responde plis

[@IntegralsForYou]

Hola Luis, el 1 lo pongo para señalar que esa función es la que vamos a poner en "dv" cuando se hace la integración por partes. No es necesario escribirlo si no se quiere.

[@LUISGONZALEZ-ir5pg]

@@IntegralsForYou gracias gracias muchas gracias enserio...entrego trabajos en media hora y me salvaste la vida bro

[@IntegralsForYou]

LUIS GONZALEZ Me alegro mucho haberte servido de ayuda! Ya me dirás qué nota obtienes! 😉

[@fabriiimg2604]

alguien me puede decir de donde sale ese +1 -1??

[@IntegralsForYou]

Hola Fabriii mussini! En primer lugar hay que entender que sumar y restar la misma cantidad, en este caso 1, no cambia el resultado porque es como si estuvieramos sumando cero. En segundo lugar, hacemos +1-1 porque como en el denominador tenemos 1+x^2 y en el numerador x^2, si le sumamos +1 tendremos lo mismo en el numerador y denominador y se podrá simplificar: (1+x^2)/(1+x^2) = 1 y esto es facil de integrar. Pero si sumamos +1, hay que restarle -1 para que la integral no cambie. Por suerte, 1/(1+x^2) también es facil de integrar. Teoricamente deberiamos hacer la division polinomial D=dq+r, pero si tenemos un poco de ojo antes y podemos utilizar el truco que te acabo de explicar, te ahorras mucho tiempo. Si haces la division polinomial, el resultado seria este: D = x^2 d = 1+x^2 q = 1 r = -1 De modo que D = dq + r ==> x^2 = (1+x^2)*1 - 1 Aunque para integrar usamos D/d = (dq + r)/d = dq/d + r/d = q + r/d ==> x^2/(1+x^2) = 1 - 1/(1+x^2) Para obtener lo mismo más rápidamente usamos el +1-1 : x^2/(1+x^2) = (x^2+1-1)/(1+x^2) = (1+x^2)/(1+x^2) - 1/(1+x^2) = 1 - 1/(1+x^2) Sea cual sea el método que uses, siempre deberías obtener el mismo resultado. Un saludo, espero haberte ayudado! ;-D

[@briiandaromeroo6000]

no entendi el resultado :(

[@IntegralsForYou]

Si me dices que no entiendes exactamente intentaré echarte una mano :-D

[@Dobzhik]

спас брат

[@IntegralsForYou]

😀

[@fatimazohra4893]

heloo i'm looking for this x/(x^2+x+1) integral

[@IntegralsForYou]

Hi Fatima! Integral of x/(x^2+x+1) dx = = (1/2)Integral of (2x)/(x^2+x+1) dx = = (1/2)Integral of (2x+1-1)/(x^2+x+1) dx = = (1/2)[ Integral of (2x+1)/(x^2+x+1) dx - Integral of 1/(x^2+x+1) dx ] = = (1/2)[ ln|x^2+x+1| - kzbin.info/www/bejne/nqDEdKCnoNVjb5Y ] = = (1/2)[ ln|x^2+x+1| - (2/sqrt(3))arctan((2x+1)/sqrt(3)) ] = = (1/2)ln|x^2+x+1| - (1/sqrt(3))arctan((2x+1)/sqrt(3)) + C :-D

[@fatimazohra4893]

thank u so much

[@IntegralsForYou]

You're welcome!

[@coolguy8234]

So easy....its child's play

[@rubico1894]

just like shitting in the toilet, but your people dont seem to do it that much, pajeet

[@Monsenstar]

Integral (1+x²)√x dx how

[@IntegralsForYou]

Integral of (1+x^2)sqrt(x) dx = = Integral of (1+x^2)x^(1/2) dx = = Integral of (x^(1/2) + x^(5/2)) dx = = x^(3/2)/(3/2) + x^(7/2)/(7/2) = = (2/3)x^(3/2) + (2/7)x^(7/2) = = (2/21)x^(3/2)(7 + 3x^2) + C Hope it helped! 😊

[@Monsenstar]

@@IntegralsForYou thank u, u are the kind human exist. But what if we use du dx ?

[@IntegralsForYou]

@@Monsenstar Hi! No need to use substitution, but if still you want to use it it would be: Integral of (1+x^2)sqrt(x) dx = Substitution: u = sqrt(x) ==> u^2 = x ==> u^4 = x^2 du = 1/2sqrt(x) dx = 1/2u dx ==> 2u du = dx = Integral of (1+u^4)u 2u du = = 2*Integral of (u^2 + u^6) du = = 2*(u^3/3 + u^7/7) = = 2*((1/3)u^3 + (1/7)u^7) = = (2/21)*(u^3)(7 + 3u^4) = = (2/21)*((sqrt(x)^3))(7 + 3x^2) = = (2/21)x^(3/2)(7 + 3x^2) + C :-D

[@shobhaverma1223]

From which country you belong😀?

[@IntegralsForYou]

Hi! I'm from Spain 😉

[@shobhaverma1223]

@@IntegralsForYou Lovely country♥️

[@IntegralsForYou]

@Shreya Verma 😊

[@celerinorodriguezgonzalez3508]

consejo agregar explicacion oral

[@IntegralsForYou]

Hola Mat! Gracias por el consejo! Ya me lo planteé al principio, pero decidí que prefiero que penséis vosotros por qué hago cada paso y si hay dudas las aclaramos en los comentarios. En mi opinión es una buena técnica para aprender 😊

[@celerinorodriguezgonzalez3508]

@@IntegralsForYou Buen punto, solo que seria bueno de apoyo para los que aun no comprenden algunas reglas o que no conocen los métodos de integración por ejemplo.

[@IntegralsForYou]

@@celerinorodriguezgonzalez3508 Sí, cierto! Habrá gente que este método le convenga y otra que no... Al final cada uno tiene su mejor manera de aprender. Agradezco mucho tu aportación, Mat. Un saludo!

[@user-rc8fm3ue2m]

махталитет )

[@jadrienhannon1625]

What's the point of integrating by parts if you just make u your original equation anyways? Lol. u should be 1 + x^2 in this case. This is so pointlessly difficult.

[@IntegralsForYou]

Hi! "u" cannot be 1+x^2 because we need two functions multiplying themselves and ln(1+x^2) is not a multiplication. We have the same problem with the integral of ln(x) 👉 kzbin.info/www/bejne/l2KQXnhoqrSneJo . The only way to solve the integral of ln(x) and the integral of ln(1+x^2) by parts is by doing u=ln(1+x^2) and dv = 1dx

[@oussamabenbrahim4848]

It's wrong if you're integration is between infinite

[@IntegralsForYou]

Hi! In infinite the integral diverges, yes

[@alihaydarsavasc4200]

thank you you saw greath

[@IntegralsForYou]

You're welcome! :-D

[@zullyzulema4605]

dudeeeeeee wtf? where is your voice?it would be a great video if you just could explain a little bit. Add value to your videos...

[@johnrachidy5252]

honestly , It's totally understable without his voice ..

[@akele02]

Zully Zulema nuh if you read your lessons you will understand easily

[@celerinorodriguezgonzalez3508]

l

[@IntegralsForYou]

O

[@pabloogl_]

WHAT THE FUCK!!

[@IntegralsForYou]

A positive WTF or a negative WTF? 😜

[@Icekilling]

smurf.... why smurfing

[@jadrienhannon1625]

The worst way to do this ever

[@IntegralsForYou]

I didn't know there are other ways to do it...! Can you share them, please?